Just as the binomial coefficient \(\dbinom{n}{r}=\dbinom{n}{r,n-r}\) is the coefficient of \(x^ry^{n-r}\) in the expansion of the binomial expression \((x+y)^n\), the multinomial coefficient \(\dbinom{9}{4,3,2}\) is the coefficient of \(x^4y^3z^2\) in the expansion of \((x+y+z)^9\).

The general multinomial coefficient \(\dbinom{n}{k_1, k_2, \dotsc, k_s}\) is likewise the coefficient of \(x_1^{k_1}x_2^{k_2}\dotsm x_s^{k_s}\) in the expansion of \((x_1+x_2+\dotsb+x_s)^n\).

A bag contains six white and four black marbles. Find the chance that, if two marbles are drawn together, they are both black.

Approach 1:

Let us consider the marbles to be labelled.

Then for drawing two marbles together we have \(\dbinom{10}{2}\) possibilities.

Also, there are \(\dbinom{4}{2}\) possible ways to draw two black marbles, as there are four black marbles in total.

Therefore the chance that, when drawing two marbles, they are both black is \[\frac{\dbinom{4}{2}}{\dbinom{10}{2}} = \frac{6}{45} = \frac{2}{15}.\]

Approach 2:

Notice that drawing two marbles at the same time is the same as drawing two marbles consecutively without replacing the first marble.

Drawing the first marble, we have a chance (probability) of \(\dfrac{4}{10}=\dfrac{2}{5}\) for it to be black, as there are four black marbles and ten marbles in total.

For the second marble, we have a chance of \(\dfrac{3}{9} = \dfrac{1}{3}\) that it’s black, as there are nine marbles left, and three of them are black.

Therefore, in total we have a chance of \[\frac{2}{5} \times \frac{1}{3} = \frac{2}{15}\] for both marbles to be black.